2018A AMC 12 真题（P版）

1 答案 D

原来有 \(36\) 个红球，而且这个数量保持不变。若红球要占全瓮的 \(72\%\)，瓮中必须有 \(36 \div 0.72 = 50\) 个球。因为 \(100 - 50 = 50\)，所以正好要取出 \(50\) 个蓝球。

所以正确答案是 D。

English

There are \(36\) red balls, and this count stays fixed. For the red balls to be \(72\%\) of the urn, the urn must contain \(36 \div 0.72 = 50\) balls. Since \(100 - 50 = 50\), exactly \(50\) blue balls are removed.

Thus, the correct answer is D.

2 答案 C

三种石头每磅分别价值 $\(2.80\)、$\(2.75\) 和 $\(2\)。对每一种可能的 \(5\) 磅石头数量，尽量装入 \(4\) 磅石头，再用 \(1\) 磅石头填满剩余容量。取 \(0,1,2\) 或 \(3\) 块 \(5\) 磅石头时，最大价值分别为 $\(48\)、$\(49\)、$\(50\) 和 $\(48\)。

因此，两块 \(5\) 磅石头和两块 \(4\) 磅石头恰好用满 \(18\) 磅，最大价值是 $\(50\)。所以正确答案是 C。

English

The rocks are worth $\(2.80,\) $\(2.75,\) and $\(2\) per pound, respectively. For each possible number of \(5\)-pound rocks, use as many \(4\)-pound rocks as fit and fill any leftover capacity with \(1\)-pound rocks. Taking \(0,1,2,\) or \(3\) of the \(5\)-pound rocks gives maximum values $\(48,\) $\(49,\) $\(50,\) and $\(48,\) respectively.

Thus two \(5\)-pound and two \(4\)-pound rocks use all \(18\) pounds, and the maximum value is $\(50.\) Thus, the correct answer is C.

3 答案 E

三个不相邻课时的选择为 \(\{1,3,5\}\)，\(\{1,3,6\}\)，\(\{1,4,6\}\)，和 \(\{2,4,6\}\)，共 \(4\) 种。三门不同课程可以按任意顺序放进这样的课时集合中，有 \(3! = 6\) 种顺序，因此共有 \(4 \cdot 6 = 24\) 种安排。

所以正确答案是 E。

English

The choices of three non-consecutive periods are \(\{1,3,5\},\) \(\{1,3,6\},\) \(\{1,4,6\},\) and \(\{2,4,6\},\) a total of \(4.\) The three distinct courses can be placed into any such set in \(3! = 6\) orders, giving \(4 \cdot 6 = 24\) schedules.

Thus, the correct answer is E.

4 答案 D

否定这三句假话，得到 \(d \lt 6\)，\(d \gt 5\)，和 \(d \gt 4\)。这些条件的交集是 \(5 \lt d \lt 6\)，也就是区间 \((5, 6)\)。

所以正确答案是 D。

English

Negating the three false statements gives \(d \lt 6,\) \(d \gt 5,\) and \(d \gt 4.\) The intersection of these conditions is \(5 \lt d \lt 6,\) that is, the interval \((5, 6).\)

Thus, the correct answer is D.

5 答案 E

因为 \(x^2 - 3x + 2 = (x-1)(x-2)\)，它的根是 \(1\) 和 \(2\)。如果 \(1\) 是公共根，则 \(1 - 5 + k = 0\)，所以 \(k = 4\)。如果 \(2\) 是公共根，则 \(4 - 10 + k = 0\)，所以 \(k = 6\)。可能值之和为 \(4 + 6 = 10\)。

所以正确答案是 E。

English

Since \(x^2 - 3x + 2 = (x-1)(x-2),\) its roots are \(1\) and \(2.\) If \(1\) is a shared root then \(1 - 5 + k = 0,\) so \(k = 4.\) If \(2\) is a shared root then \(4 - 10 + k = 0,\) so \(k = 6.\) The sum of possible values is \(4 + 6 = 10.\)

Thus, the correct answer is E.

6 答案 B

因为 \(m + 10 \lt n + 1\)，这六个数已经按递增顺序排列，所以中位数是中间两个数的平均：\(\frac{(m+10)+(n+1)}{2} = n\)，得到 \(m = n - 11\)。平均数条件为 \[ \tiny \frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n\text{，} \] 因此 \(7n - 16 = 6n\)，得 \(n = 16\)。于是 \(m = 5\)，所以 \(m + n = 21\)。

所以正确答案是 B。

English

Because \(m + 10 \lt n + 1,\) the six numbers are already increasing, so the median is the average of the middle two: \(\frac{(m+10)+(n+1)}{2} = n,\) giving \(m = n - 11.\) The mean condition is \[ \tiny \frac{(n-11)+(n-7)+(n-1)+(n+1)+(n+2)+2n}{6} = n, \] so \(7n - 16 = 6n\) and \(n = 16.\) Then \(m = 5,\) and \(m + n = 21.\)

Thus, the correct answer is B.

7 答案 E

因为 \(4000 = 2^5 \cdot 5^3\)，原式等于 \(2^{5+n} \cdot 5^{3-n}\)。它是整数当且仅当 \(5 + n \ge 0\) 且 \(3 - n \ge 0\)，也就是 \(-5 \le n \le 3\)。这样的整数有 \(3 - (-5) + 1 = 9\) 个。

所以正确答案是 E。

English

Since \(4000 = 2^5 \cdot 5^3,\) the expression equals \(2^{5+n} \cdot 5^{3-n}.\) This is an integer exactly when both \(5 + n \ge 0\) and \(3 - n \ge 0,\) that is, \(-5 \le n \le 3.\) There are \(3 - (-5) + 1 = 9\) such integers.

Thus, the correct answer is E.

8 答案 E

底边 \(DE\) 所属的 \(\triangle ADE\) 与最小三角形相似，且底边是其 \(4\) 倍，所以根据相似图形面积的平方缩放，\([ADE] = 4^2 \cdot 1 = 16\)。梯形 \(DBCE\) 是 \(\triangle ABC\) 中剩下的部分，所以它的面积是 \(40 - 16 = 24\)。

所以正确答案是 E。

English

The base \(DE\) of \(\triangle ADE\) is \(4\) times the base of a smallest triangle, so by the square scaling of similar areas, \([ADE] = 4^2 \cdot 1 = 16.\) The trapezoid \(DBCE\) is what remains of \(\triangle ABC,\) so its area is \(40 - 16 = 24.\)

Thus, the correct answer is E.

9 答案 E

对于 \(0 \le x \le \pi\) 和 \(0 \le y \le \pi\)，有 \(\sin x \ge 0\)、\(\sin y \ge 0\)、\(\cos x \le 1\) 且 \(\cos y \le 1\)。因此 \[ \begin{aligned} \sin(x+y) &= \sin x \cos y \\ &\quad {}+ \cos x \sin y \\ &\le \sin x + \sin y\text{。} \end{aligned} \] 所以这个不等式对每个满足 \(0 \le y \le \pi\) 的 \(y\) 都成立。

所以正确答案是 E。

English

For \(0 \le x \le \pi\) and \(0 \le y \le \pi\) we have \(\sin x \ge 0,\) \(\sin y \ge 0,\) \(\cos x \le 1,\) and \(\cos y \le 1.\) Hence \[ \begin{aligned} \sin(x+y) &= \sin x \cos y \\ &\quad {}+ \cos x \sin y \\ &\le \sin x + \sin y. \end{aligned} \] The inequality therefore holds for every \(y\) with \(0 \le y \le \pi.\)

Thus, the correct answer is E.

10 答案 C

第二个方程给出 \(|x| - |y| = \pm 1\)，等价于 \(x = \pm y \pm 1\)。代入 \(x + 3y = 3\)：

若 \(x = y + 1\) 则 \((x, y) = \left(\tfrac32, \tfrac12\right)\)。若 \(x = y - 1\) 则 \((x, y) = (0, 1)\)。若 \(x = -y + 1\)，又得到 \((x, y) = (0, 1)\)。若 \(x = -y - 1\) 则 \((x, y) = (-3, 2)\)。

不同的解是 \((-3, 2)\)，\((0, 1)\)，和 \(\left(\tfrac32, \tfrac12\right)\)，它们都满足原方程，所以共有 \(3\) 个。

所以正确答案是 C。

English

The second equation gives \(|x| - |y| = \pm 1,\) equivalently \(x = \pm y \pm 1.\) Substituting into \(x + 3y = 3:\)

If \(x = y + 1,\) then \((x, y) = \left(\tfrac32, \tfrac12\right).\) If \(x = y - 1,\) then \((x, y) = (0, 1).\) If \(x = -y + 1,\) then again \((x, y) = (0, 1).\) If \(x = -y - 1,\) then \((x, y) = (-3, 2).\)

The distinct solutions are \((-3, 2),\) \((0, 1),\) and \(\left(\tfrac32, \tfrac12\right),\) all of which check, so there are \(3.\)

Thus, the correct answer is C.

11 答案 D

折痕位于 \(AB\) 的垂直平分线上，并且由于 \(AC \gt BC\)，它在 \(E\) 处与 \(AC\) 相交。设 \(D\) 为 \(AB\) 的中点，则 \(AD = \tfrac52\)，且 \(\triangle ADE\) 在 \(D\) 处为直角。因为 \(\triangle ADE \sim \triangle ACB\)，有 \(\tfrac{DE}{AD} = \tfrac{CB}{AC} = \tfrac34\)，所以 \[ DE = \frac52 \cdot \frac34 = \frac{15}{8}\text{。}\]

所以正确答案是 D。

English

The crease lies along the perpendicular bisector of \(AB,\) meeting \(AC\) at \(E\) because \(AC \gt BC.\) Let \(D\) be the midpoint of \(AB,\) so \(AD = \tfrac52\) and \(\triangle ADE\) is right-angled at \(D.\) Since \(\triangle ADE \sim \triangle ACB,\) we have \(\tfrac{DE}{AD} = \tfrac{CB}{AC} = \tfrac34,\) so \[ DE = \frac52 \cdot \frac34 = \frac{15}{8}. \]

Thus, the correct answer is D.

12 答案 C

将 \(\{1, \ldots, 12\}\) 分成六条整除链 \(\{1,2,4,8\}\)，\(\{3,6,12\}\)，\(\{5,10\}\)，\(\{7\}\)，\(\{9\}\)，\(\{11\}\)。因为 \(S\) 中不能有一个元素整除另一个元素，所以每条链至多贡献一个元素；需要 \(6\) 个元素就迫使每条链正好取一个，因此 \(7, 9, 11 \in S\)。

因为 \(9 \in S\)，所以 \(3 \notin S\)，第二条链贡献 \(6\) 或 \(12\)，于是第一条链中既不能选 \(1\) 也不能选 \(2\)（它们会整除 \(6\) 和 \(12\)）。从第一条链取 \(4\) 可以做到：\(S = \{4, 5, 6, 7, 9, 11\}\) 满足条件。因此最小可能元素是 \(4\)。

所以正确答案是 C。

English

Partition \(\{1, \ldots, 12\}\) into the six divisibility chains \(\{1,2,4,8\},\) \(\{3,6,12\},\) \(\{5,10\},\) \(\{7\},\) \(\{9\},\) \(\{11\}.\) Since no element of \(S\) may divide another, at most one comes from each chain; needing \(6\) elements forces exactly one from each, so \(7, 9, 11 \in S.\)

Because \(9 \in S,\) \(3 \notin S,\) so the second chain contributes \(6\) or \(12,\) and then neither \(1\) nor \(2\) can be chosen from the first chain (they divide \(6\) and \(12\)). Taking \(4\) from the first chain works: \(S = \{4, 5, 6, 7, 9, 11\}\) has the property. Hence the least possible element is \(4.\)

Thus, the correct answer is C.

13 答案 D

给每个 \(a_i\) 都加 \(1\)，可在这些表达式与从 \(0\) 到 \(3^8 - 1\) 的 \(3\) 进制数之间建立双射，所以恰有 \(3^8 = 6561\) 个不同整数出现。它们关于 \(0\) 对称（把所有 \(a_i\) 取相反数会使数值取相反数），所以除了 \(0\) 本身外，一半为正。非负整数的个数为 \[ 1 + \tfrac12(6561 - 1) = 3281 \] 即从 \(0\) 到 \(3280\) 的所有整数。

所以正确答案是 D。

English

Adding \(1\) to every \(a_i\) gives a bijection between these expressions and the base-\(3\) numerals for \(0\) through \(3^8 - 1,\) so exactly \(3^8 = 6561\) distinct integers occur. They are symmetric about \(0\) (negating all \(a_i\) negates the value), so besides \(0\) itself, half are positive: \[ 1 + \tfrac12(6561 - 1) = 3281 \] nonnegative integers, namely \(0\) through \(3280.\)

Thus, the correct answer is D.

14 答案 D

把两个对数都写成以 \(2\) 为底：\(\tfrac{2}{\log_2 3x} = \tfrac{3}{\log_2 2x}\)，所以 \(2 \log_2 2x = 3 \log_2 3x\)，即 \((2x)^2 = (3x)^3\)。于是 \(4x^2 = 27x^3\)，得 \(x = \tfrac{4}{27}\)。因为 \(\gcd(4, 27) = 1\)，所以 \(p + q = 4 + 27 = 31\)。

所以正确答案是 D。

English

Writing both logarithms in base \(2:\) \(\tfrac{2}{\log_2 3x} = \tfrac{3}{\log_2 2x},\) so \(2 \log_2 2x = 3 \log_2 3x,\) i.e. \((2x)^2 = (3x)^3.\) Then \(4x^2 = 27x^3,\) giving \(x = \tfrac{4}{27}.\) Since \(\gcd(4, 27) = 1,\) we get \(p + q = 4 + 27 = 31.\)

Thus, the correct answer is D.

15 答案 B

在正方形的对称群作用下，\(49\) 个方格分成若干轨道，同一轨道中的每个方格必须颜色相同。以中心为原点，给方格坐标 \((i,j)\)，其中 \(-3\le i,j\le3\)。旋转和翻折可以改变坐标符号并交换两个坐标，所以每个轨道都有唯一代表满足 \(0\le i\le j\le3\)。这样的数对共有 \(1+2+3+4=10\) 个。每个轨道可选黑色或白色，得到 \(2^{10}\) 种着色，但要排除全黑和全白的网格。因此共有 \(2^{10} - 2 = 1022\) 个对称扫描码。

因此，正确答案是 B。

English

Under the symmetry group of the square, the \(49\) cells break into orbits, and every cell in an orbit must have the same color. Give a cell coordinates \((i,j)\) relative to the center, where \(-3\le i,j\le3.\) Rotations and reflections can change signs and interchange the coordinates, so each orbit has one representative with \(0\le i\le j\le3.\) There are \(1+2+3+4=10\) such pairs. Each orbit is black or white, giving \(2^{10}\) colorings, but the all-black and all-white grids are excluded. So there are \(2^{10} - 2 = 1022\) symmetric scanning codes.

Thus, the correct answer is B.

16 答案 E

将 \(x^2 = y + a\) 代入 \(x^2 + y^2 = a^2\)，得 \(y^2 + y + (a - a^2) = 0\)，可分解为 \((y + 1 - a)(y + a) = 0\)，所以 \(y = a - 1\) 或 \(y = -a\)。它们分别对应 \(x^2 = 2a - 1\) 和 \(x^2 = 0\)。

方程 \(x^2 = 0\) 总是给出单个点 \((0, -a)\)，即抛物线的顶点。方程 \(x^2 = 2a - 1\) 恰好在 \(2a - 1 \gt 0\)，即 \(a \gt \tfrac12\) 时给出另外两个点。因此恰好有 \(3\) 个交点当且仅当 \(a \gt \tfrac12\)。

所以正确答案是 E。

English

Substituting \(x^2 = y + a\) into \(x^2 + y^2 = a^2\) gives \(y^2 + y + (a - a^2) = 0,\) which factors as \((y + 1 - a)(y + a) = 0,\) so \(y = a - 1\) or \(y = -a.\) These correspond to \(x^2 = 2a - 1\) and \(x^2 = 0.\)

The equation \(x^2 = 0\) always gives the single point \((0, -a),\) the vertex of the parabola. The equation \(x^2 = 2a - 1\) gives two more points exactly when \(2a - 1 \gt 0,\) i.e. \(a \gt \tfrac12.\) So there are \(3\) intersection points precisely when \(a \gt \tfrac12.\)

Thus, the correct answer is E.

17 答案 D

将直角放在原点，两条直角边沿坐标轴，则顶点为 \((4, 0)\)、\((0, 3)\)、\((0, 0)\)，正方形 \(S\) 为 \([0, s] \times [0, s]\)。斜边是 \(3x + 4y - 12 = 0\)，它到正方形最近的角 \((s, s)\) 的距离满足 \[ \frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2\text{。}\] 这给出 \(s = \tfrac{22}{7}\) 或 \(s = \tfrac27\)；只有 \(s = \tfrac27\) 能使正方形留在三角形内部。

田地面积为 \(\tfrac12 \cdot 3 \cdot 4 = 6\)，未种植正方形面积为 \(\left(\tfrac27\right)^2 = \tfrac{4}{49}\)。 已种植比例为 \[ 1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \frac{145}{147}\text{。}\]

所以正确答案是 D。

English

Place the right angle at the origin with legs on the axes, so the vertices are \((4, 0),\) \((0, 3),\) \((0, 0),\) and the square \(S\) is \([0, s] \times [0, s].\) The hypotenuse is \(3x + 4y - 12 = 0,\) and the distance from its nearest corner \((s, s)\) is \[ \frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2. \] This gives \(s = \tfrac{22}{7}\) or \(s = \tfrac27;\) only \(s = \tfrac27\) keeps the square inside the triangle.

The field has area \(\tfrac12 \cdot 3 \cdot 4 = 6\) and the unplanted square has area \(\left(\tfrac27\right)^2 = \tfrac{4}{49}.\) The planted fraction is \[ 1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \frac{145}{147}. \]

Thus, the correct answer is D.

18 答案 D

因为 \(D\) 和 \(E\) 是中点，\(\triangle ADE\) 的面积是 \(\triangle ABC\) 面积的 \(\tfrac14\) 即 \(30\)，所以梯形 \(EDBC\) 的面积为 \(120 - 30 = 90\)。

由角平分线定理，\(G\) 将 \(BC\) 分成 \(BG = \tfrac{AB}{AB + AC} \cdot BC = \tfrac56 BC\)，同样 \(F\) 将 \(DE\) 分成 \(DF = \tfrac56 DE\)。因为 \(FDBG\) 和 \(EDBC\) 高相同，\(FDBG\) 的面积是 \(EDBC\) 面积的 \(\tfrac56\)，即 \(\tfrac56 \cdot 90 = 75\)。

所以正确答案是 D。

English

Since \(D\) and \(E\) are midpoints, \(\triangle ADE\) has \(\tfrac14\) the area of \(\triangle ABC,\) namely \(30,\) so trapezoid \(EDBC\) has area \(120 - 30 = 90.\)

By the Angle Bisector Theorem, \(G\) divides \(BC\) with \(BG = \tfrac{AB}{AB + AC} \cdot BC = \tfrac56 BC,\) and likewise \(F\) divides \(DE\) so that \(DF = \tfrac56 DE.\) Because \(FDBG\) and \(EDBC\) share the same height, the area of \(FDBG\) is \(\tfrac56\) of the area of \(EDBC:\) \(\tfrac56 \cdot 90 = 75.\)

Thus, the correct answer is D.

19 答案 C

\(A\) 中每个元素都可唯一写成 \(2^i 3^j 5^k\)，其中 \(i, j, k \ge 0\)，所以所有倒数的和可分解为 \[ \begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}\text{。} \end{aligned} \] 这等于 \(2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}\)。因为 \(\gcd(15, 4) = 1\)，所以 \(m + n = 15 + 4 = 19\)。

所以正确答案是 C。

English

Each element of \(A\) is uniquely \(2^i 3^j 5^k\) with \(i, j, k \ge 0,\) so summing all reciprocals factors as \[ \begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}. \end{aligned} \] This equals \(2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.\) With \(\gcd(15, 4) = 1,\) \(m + n = 15 + 4 = 19.\)

Thus, the correct answer is C.

20 答案 D

因为 \(\triangle ABC\) 是等腰直角三角形，\(CM = BM = \tfrac32 \sqrt2\)，且 \(B, C\) 处的底角为 \(45^\circ\)。由于 \(AIME\) 是圆内接四边形且 \(A\) 处是直角，角 \(\angle IME = 90^\circ\)。令 \(x = CI\)，\(y = BE\)。在 \(\triangle MCI\) 中由余弦定理， \[ \begin{aligned} IM^2 &= x^2 + \tfrac92 \\ &\quad {}- 2 \cdot x \cdot \tfrac32\sqrt2 \cdot \cos 45^\circ \\ &= x^2 - 3x + \tfrac92\text{，} \end{aligned} \] 同理，\(ME^2 = y^2 - 3y + \tfrac92\)。

对直角三角形 \(EMI\) 和 \(IAE\) 使用勾股定理，得到 \(IM^2 + ME^2\) \(= (3-x)^2\) \(+ (3-y)^2\)，化简为 \(x + y = 3\)。面积条件 \(\tfrac12 IM \cdot ME = 2\) 表示 \(IM^2 \cdot ME^2 = 16\)。代入 \(y = 3 - x\) 后，\(ME^2 = x^2 - 3x + \tfrac92 = IM^2\)，所以 \(\left(x^2 - 3x + \tfrac92\right)^2 = 16\)，因而 \(x^2 - 3x + \tfrac92 = 4\)，即 \(x^2 - 3x + \tfrac12 = 0\)。

因为 \(AI \gt AE\) 迫使 \(y \gt x\)， 所以取较小根 \(x = \tfrac{3 - \sqrt{7}}{2}\)。 于是 \(a + b + c = 3 + 7 + 2 = 12\)。

所以正确答案是 D。

English

Since \(\triangle ABC\) is an isosceles right triangle, \(CM = BM = \tfrac32 \sqrt2\) and the base angles at \(B, C\) are \(45^\circ.\) As \(AIME\) is cyclic with right angle at \(A,\) angle \(\angle IME = 90^\circ.\) Let \(x = CI\) and \(y = BE.\) By the Law of Cosines in \(\triangle MCI,\) \[ \begin{aligned} IM^2 &= x^2 + \tfrac92 \\ &\quad {}- 2 \cdot x \cdot \tfrac32\sqrt2 \cdot \cos 45^\circ \\ &= x^2 - 3x + \tfrac92, \end{aligned} \] and similarly \(ME^2 = y^2 - 3y + \tfrac92.\)

The Pythagorean Theorem in right triangles \(EMI\) and \(IAE\) gives \(IM^2 + ME^2\) \(= (3-x)^2\) \(+ (3-y)^2,\) which simplifies to \(x + y = 3.\) The area condition \(\tfrac12 IM \cdot ME = 2\) means \(IM^2 \cdot ME^2 = 16.\) Substituting \(y = 3 - x\) makes \(ME^2 = x^2 - 3x + \tfrac92 = IM^2,\) so \(\left(x^2 - 3x + \tfrac92\right)^2 = 16,\) hence \(x^2 - 3x + \tfrac92 = 4,\) i.e. \(x^2 - 3x + \tfrac12 = 0.\)

Since \(AI \gt AE\) forces \(y \gt x,\) we take the smaller root \(x = \tfrac{3 - \sqrt{7}}{2}.\) Then \(a + b + c = 3 + 7 + 2 = 12.\)

Thus, the correct answer is D.

21 答案 B

选项 A-D 中每个多项式都没有正根，且恰有一个负根，位于 \((-1, 0)\)（在 \(0\) 处为正，在 \(-1\) 处为负），并且在该区间递增。在区间 \((-1, 0)\) 上，\(x^{19} \gt x^{17}\) 且 \(x^{13} \gt x^{11}\)。因此在这个区间的每一点上，A、C、D 的取值都大于 B，于是它们与 \(0\) 的交点都比 B 更靠左。所以在 A 到 D 中，B 的根最大。

一次的选项 E 的根为 \(e=-\tfrac{2018}{2019}\)。因为 \(\tfrac{2018}{2019}\gt\tfrac9{10}\)，所以 \(e^{17}+2018e^{11}+1\) \(\lt-2018\left(\tfrac9{10}\right)^{11}+1\) \(\lt0\)。由于选项 B 的多项式单调递增，且在 \(e\) 处取负值，它的根在 E 的根的右侧。因此 B 的实根最大。

所以正确答案是 B。

English

Each polynomial in choices A–D has no positive root and exactly one negative root, which lies in \((-1, 0)\) (it is positive at \(0\) and negative at \(-1\)) and is increasing there. On the interval \((-1, 0),\) \(x^{19} \gt x^{17}\) and \(x^{13} \gt x^{11}.\) Thus each of A, C, and D has a larger value than B at every point of this interval, so each crosses \(0\) to the left of B. Therefore B has the greatest root among A–D.

The linear choice E has root \(e=-\tfrac{2018}{2019}.\) Since \(\tfrac{2018}{2019}\gt\tfrac9{10},\) we have \(e^{17}+2018e^{11}+1\) \(\lt-2018\left(\tfrac9{10}\right)^{11}+1\) \(\lt0.\) Because B is increasing and is negative at \(e,\) its root lies to the right of E’s. Hence B has the greatest real root.

Thus, the correct answer is B.

22 答案 A

写 \(z = a + bi\)，由 \((a + bi)^2 = 4 + 4\sqrt{15}\,i\) 得 \(a^2 - b^2 = 4\) 和 \(2ab = 4\sqrt{15}\)。因此 \(a^4 - 4a^2 - 60 = 0\)，即 \((a^2 - 10)(a^2 + 6) = 0\)，得 \(a = \pm\sqrt{10}\)，\(b = \pm\sqrt{6}\)。第一个方程给出的顶点为 \(\pm(\sqrt{10} + \sqrt{6}\,i)\)。对 \(z^2 = 2 + 2\sqrt{3}\,i\) 用同样方法得到 \(\pm(\sqrt{3} + i)\)。

对 \((\sqrt{10}, \sqrt{6})\)，\((\sqrt{3}, 1)\)，\((-\sqrt{10}, -\sqrt{6})\)，\((-\sqrt{3}, -1)\) 使用鞋带公式，面积为 \(6\sqrt{2} - 2\sqrt{10}\)。因此 \(p + q + r + s\) \(= 6 + 2 + 2 + 10\) \(= 20\)。

所以正确答案是 A。

English

Writing \(z = a + bi\) with \((a + bi)^2 = 4 + 4\sqrt{15}\,i\) gives \(a^2 - b^2 = 4\) and \(2ab = 4\sqrt{15}.\) Then \(a^4 - 4a^2 - 60 = 0,\) so \((a^2 - 10)(a^2 + 6) = 0,\) yielding \(a = \pm\sqrt{10},\) \(b = \pm\sqrt{6}.\) The vertices from the first equation are \(\pm(\sqrt{10} + \sqrt{6}\,i).\) The same method on \(z^2 = 2 + 2\sqrt{3}\,i\) gives \(\pm(\sqrt{3} + i).\)

Applying the shoelace formula to \((\sqrt{10}, \sqrt{6}),\) \((\sqrt{3}, 1),\) \((-\sqrt{10}, -\sqrt{6}),\) \((-\sqrt{3}, -1)\) gives area \(6\sqrt{2} - 2\sqrt{10}.\) Thus \(p + q + r + s\) \(= 6 + 2 + 2 + 10\) \(= 20.\)

Thus, the correct answer is A.

23 答案 E

将 \(PN\) 经过 \(N\) 延长到 \(Q\)，使 \(PN = NQ\)。 因为 \(N\) 是 \(UG\) 和 \(PQ\) 的中点，四边形 \(UPGQ\) 是平行四边形，所以 \(GQ \parallel PT\) 且 \(GQ = PU = 1 = AG\)。 因此 \(\angle QGA\) \(= 180^\circ - \angle T\) \(= \angle P + \angle A\) \(= 36^\circ + 56^\circ = 92^\circ\)， 等腰三角形 \(QGA\) 给出 \(\angle QAG = \tfrac12(180^\circ - 92^\circ) = 44^\circ\)。

因为 \(M, N\) 是中点，\(MN\) 是 \(\triangle QPA\) 的中位线，所以 \(MN \parallel AQ\)，并且 \[ \begin{aligned} \angle NMP &= \angle QAP \\ &= \angle QAG + \angle GAP \\ &= 44^\circ + 56^\circ = 100^\circ\text{。} \end{aligned} \] 因此直线 \(MN\) 和 \(PA\) 所成锐角为 \(180^\circ - 100^\circ = 80^\circ\)。

所以正确答案是 E。

English

Extend \(PN\) through \(N\) to \(Q\) with \(PN = NQ.\) Since \(N\) is the midpoint of \(UG\) and of \(PQ,\) the quadrilateral \(UPGQ\) is a parallelogram, so \(GQ \parallel PT\) and \(GQ = PU = 1 = AG.\) Then \(\angle QGA\) \(= 180^\circ - \angle T\) \(= \angle P + \angle A\) \(= 36^\circ + 56^\circ = 92^\circ,\) and the isosceles triangle \(QGA\) gives \(\angle QAG = \tfrac12(180^\circ - 92^\circ) = 44^\circ.\)

Because \(M, N\) are midpoints, \(MN\) is a midline of \(\triangle QPA,\) so \(MN \parallel AQ\) and \[ \begin{aligned} \angle NMP &= \angle QAP \\ &= \angle QAG + \angle GAP \\ &= 44^\circ + 56^\circ = 100^\circ. \end{aligned} \] The acute angle between line \(MN\) and \(PA\) is therefore \(180^\circ - 100^\circ = 80^\circ.\)

Thus, the correct answer is E.

24 答案 B

若 \(c \le \tfrac12\)，Carol 一定小于 Bob，所以她只有在 Alice 小于 \(c\) 时获胜，概率为 \(c \le \tfrac12\)。若 \(c \ge \tfrac23\)，她获胜的概率为 \(1 - c \le \tfrac13\)。这两种情况都不超过 \(\tfrac12\)。

对于 \(\tfrac12 \lt c \lt \tfrac23\)，Bob 的数大于 \(c\) 的概率是 \(\frac{\frac{2}{3} - c}{\frac{2}{3} - \frac{1}{2}} = 4 - 6c\)，所以 Carol 大于 Alice 且小于 Bob 的概率是 \(c(4 - 6c)\)；反向排序的概率是 \((1 - c)(6c - 3)\)。相加得\[ \begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3 \end{aligned}\text{。}\] 这个开口向下的抛物线在 \(c = \tfrac{13}{24}\) 处取到最大值，该点位于 \(\left(\tfrac12, \tfrac23\right)\) 内，且最大值超过 \(\tfrac12\)。

所以正确答案是 B。

English

If \(c \le \tfrac12,\) Carol beats Bob automatically, so she wins only if Alice is below \(c,\) probability \(c \le \tfrac12.\) If \(c \ge \tfrac23,\) she wins with probability \(1 - c \le \tfrac13.\) Neither case exceeds \(\tfrac12.\)

For \(\tfrac12 \lt c \lt \tfrac23,\) the chance Bob’s number exceeds \(c\) is \(\frac{\frac{2}{3} - c}{\frac{2}{3} - \frac{1}{2}} = 4 - 6c,\) so the probability Carol is above Alice and below Bob is \(c(4 - 6c);\) the reverse ordering has probability \((1 - c)(6c - 3).\) Adding, \[ \begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3. \end{aligned} \] This downward parabola is maximized at \(c = \tfrac{13}{24},\) which lies in \(\left(\tfrac12, \tfrac23\right),\) and its value exceeds \(\tfrac12.\)

Thus, the correct answer is B.

25 答案 D

利用 \(A_n = a \cdot \tfrac{10^n - 1}{9}\)、\(B_n = b \cdot \tfrac{10^n - 1}{9}\) 以及 \(C_n = c \cdot \tfrac{10^{2n} - 1}{9}\)，方程 \(C_n - B_n = A_n^2\) 在除以 \(10^n - 1\) 并清分母后变为 \[ (9c - a^2) \cdot 10^n = 9b - 9c - a^2\text{。} \] 若它要对两个不同的 \(n\) 成立，则 \(10^n\) 的系数必须为零，所以 \(9c = a^2\)，进而 \(9b - 9c - a^2 = 0\)。

于是 \(c = \tfrac{a^2}{9}\) 且 \(b = 2c\)。 因此 \(a \in \{3, 6, 9\}\)，对应 \(c \in \{1, 4, 9\}\) 和 \(b \in \{2, 8, 18\}\)； 情况 \(b = 18\) 不是一个数字。有效三元组为 \((a, b, c) = (3, 2, 1)\) 和 \((6, 8, 4)\)， 而且确实有 \(4444 - 88 = 4356 = 66^2\)。 较大的数字和为 \(6 + 8 + 4 = 18\)。

所以正确答案是 D。

English

Using \(A_n = a \cdot \tfrac{10^n - 1}{9},\) \(B_n = b \cdot \tfrac{10^n - 1}{9},\) and \(C_n = c \cdot \tfrac{10^{2n} - 1}{9},\) the equation \(C_n - B_n = A_n^2\) becomes, after dividing by \(10^n - 1\) and clearing fractions, \[ (9c - a^2) \cdot 10^n = 9b - 9c - a^2. \] For this to hold at two different \(n,\) the coefficient of \(10^n\) must be zero, so \(9c = a^2\) and hence \(9b - 9c - a^2 = 0.\)

Then \(c = \tfrac{a^2}{9}\) and \(b = 2c.\) So \(a \in \{3, 6, 9\}\) with \(c \in \{1, 4, 9\}\) and \(b \in \{2, 8, 18\};\) the case \(b = 18\) is not a digit. The valid triples are \((a, b, c) = (3, 2, 1)\) and \((6, 8, 4),\) and indeed \(4444 - 88 = 4356 = 66^2.\) The greater digit sum is \(6 + 8 + 4 = 18.\)

Thus, the correct answer is D.